MonkeyF0cker
Mean People Suck
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A friend of mine and I went to play a little Pai Gow poker the other night. My friend decides to bank and gets a 9-high Pai Gow, the lowest hand possible. "What are the odds," he says. "Not good," I reply.
For those that don't know how to play Pai Gow, it is essentially a 7 card poker game. You make two hands with those seven cards - a 5-card high hand and a 2-card low hand. The high hand must be exactly that - higher than the low hand. You need to beat the dealer with both hands in order to win your bet (minus commission). If you have the same hand as the dealer (say your low hand is A7 and the dealer's low hand is A7 as well - called a copy), you lose on that particular hand. However, the dealer must beat both your high hand and low hand in order for you to lose. Needless to say, the game creates a lot of pushes where neither you nor the dealer wins with both the high and low hands.
So, if you were dealt a hand like 6A2K4AK, you'd set your high hand to AA642 and your low hand to KK. So, your high hand would be a pair of aces and your low hand a pair of kings. That's a pretty strong hand in Pai Gow.
Onto the odds.
There are two ways to get a 9-high Pai Gow. 9876432 and 9875432. In order to find the odds of getting any variant of a 9-high Pai Gow, we'll find the odds of getting one of them and simply double it. Remember, that Pai Gow also uses one joker in the deck (which can be used as a wild for a flush or straight - if not used for a flush or straight, it's an ace). So, there are actually 53 cards in the deck.
In order to find the probability of being dealt 9876432, we simply find the number of cards (outs) in the deck divided by the number of cards remaining in the deck. So, for the first card, we'd have 28 possible cards (any 9,8,7,6,4,3, or 2) out of 53 total cards that would create a situation where a 9-high Pai Gow were possible. For the second card, we'd have 24 possible out of 52 since one of our cards is already in our hand. We find the odds for all seven "outs" and multiply them to get the probability of being dealt 9876432.
So...
Card 1: 28/53 = 0.528301887
Card 2: 24/52 = 0.461538462
Card 3: 20/51 = 0.392156863
Card 4: 16/50 = 0.32
Card 5: 12/49 = 0.244897959
Card 6: 8/48 = 0.166666667
Card 7: 4/47 = 0.085106383
We multiply all of the probabilities to get the odds of being dealt the hand, which comes to: 0.000106291
However, we have a problem. That probability includes all of the possibilities of being dealt 9876432, which includes flushes. In order to alleviate this, we turn to combinatorics. First, we find the total suit combinations for our hand. It's simply 4 (the number of suits) raised to 7 (the number of cards), which yields 16,384. Now, out of those 16,384 combinations, we need to find out how many are flushes. We turn to the choose function for this.
It reads choose k from n. We have 7 cards in total where 5 cards could possibly make a flush. So, we choose 5 from 7. 7! = 5040. 5! = 120. 2! = 2. We solve and see that there are 21 combinations. However, since this is a binomial choose function, we need to multiply this by four (the number of suits). So out of 16,384 combinations, 84 (0.5126953%) make a flush. Now, we subtract 84/16384 from 1 to get the percentage of all 9876432 variants that are not flushes. We then multiply that number (0.994873047) by the probability of the entire set (0.000106291) to get the probability of being dealt a 9876432 Pai Gow.
0.994873047 * 0.000106291 = 0.000105746
Again, since this only accounts for 9876432 Pai Gows and the calculation for 9875432 Pai Gows is identical, we multiply the above number by 2. In the end, my friend had a 0.0211492% or 4727:1 chance of being dealt a 9-high Pai Gow when he chose to bank. Tough luck.
For those that don't know how to play Pai Gow, it is essentially a 7 card poker game. You make two hands with those seven cards - a 5-card high hand and a 2-card low hand. The high hand must be exactly that - higher than the low hand. You need to beat the dealer with both hands in order to win your bet (minus commission). If you have the same hand as the dealer (say your low hand is A7 and the dealer's low hand is A7 as well - called a copy), you lose on that particular hand. However, the dealer must beat both your high hand and low hand in order for you to lose. Needless to say, the game creates a lot of pushes where neither you nor the dealer wins with both the high and low hands.
So, if you were dealt a hand like 6A2K4AK, you'd set your high hand to AA642 and your low hand to KK. So, your high hand would be a pair of aces and your low hand a pair of kings. That's a pretty strong hand in Pai Gow.
Onto the odds.
There are two ways to get a 9-high Pai Gow. 9876432 and 9875432. In order to find the odds of getting any variant of a 9-high Pai Gow, we'll find the odds of getting one of them and simply double it. Remember, that Pai Gow also uses one joker in the deck (which can be used as a wild for a flush or straight - if not used for a flush or straight, it's an ace). So, there are actually 53 cards in the deck.
In order to find the probability of being dealt 9876432, we simply find the number of cards (outs) in the deck divided by the number of cards remaining in the deck. So, for the first card, we'd have 28 possible cards (any 9,8,7,6,4,3, or 2) out of 53 total cards that would create a situation where a 9-high Pai Gow were possible. For the second card, we'd have 24 possible out of 52 since one of our cards is already in our hand. We find the odds for all seven "outs" and multiply them to get the probability of being dealt 9876432.
So...
Card 1: 28/53 = 0.528301887
Card 2: 24/52 = 0.461538462
Card 3: 20/51 = 0.392156863
Card 4: 16/50 = 0.32
Card 5: 12/49 = 0.244897959
Card 6: 8/48 = 0.166666667
Card 7: 4/47 = 0.085106383
We multiply all of the probabilities to get the odds of being dealt the hand, which comes to: 0.000106291
However, we have a problem. That probability includes all of the possibilities of being dealt 9876432, which includes flushes. In order to alleviate this, we turn to combinatorics. First, we find the total suit combinations for our hand. It's simply 4 (the number of suits) raised to 7 (the number of cards), which yields 16,384. Now, out of those 16,384 combinations, we need to find out how many are flushes. We turn to the choose function for this.
It reads choose k from n. We have 7 cards in total where 5 cards could possibly make a flush. So, we choose 5 from 7. 7! = 5040. 5! = 120. 2! = 2. We solve and see that there are 21 combinations. However, since this is a binomial choose function, we need to multiply this by four (the number of suits). So out of 16,384 combinations, 84 (0.5126953%) make a flush. Now, we subtract 84/16384 from 1 to get the percentage of all 9876432 variants that are not flushes. We then multiply that number (0.994873047) by the probability of the entire set (0.000106291) to get the probability of being dealt a 9876432 Pai Gow.
0.994873047 * 0.000106291 = 0.000105746
Again, since this only accounts for 9876432 Pai Gows and the calculation for 9875432 Pai Gows is identical, we multiply the above number by 2. In the end, my friend had a 0.0211492% or 4727:1 chance of being dealt a 9-high Pai Gow when he chose to bank. Tough luck.
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